Spring 2009 - Problem 4

Solution.

1. For convenience, we extend $f$ to $f\p{x} = f\p{1}$ for $x \geq 1$. By assumption,

   $$f'\p{x} = \liminf_{h\to0}\,\frac{f\p{x\,+ h} - f\p{x}}{h}$$

   for almost every $x$. Hence, by Fatou's lemma and a translation change of variables,

   $$\begin{aligned} \int_0^1 f'\p{x} \,\diff{x} &\leq \liminf_{h\to0}\,\int_0^1 \frac{f\p{x + h} - f\p{x}}{h} \,\diff{x} \\ &= \liminf_{h\to0}\,\frac{1}{h} \p{\int_0^1 f\p{x + h} \,\diff{x} - \int_0^1 f\p{x} \,\diff{x}} \\ &= \liminf_{h\to0}\,\frac{1}{h} \p{\int_h^{1+h} f\p{x} \,\diff{x} - \int_0^1 f\p{x} \,\diff{x}} \\ &= \liminf_{h\to0}\,\frac{1}{h} \p{\int_1^{1+h} f\p{x} \,\diff{x} + \int_h^1 f\p{x} \,\diff{x} - \int_0^1 f\p{x} \,\diff{x}} \\ &= \liminf_{h\to0}\,\frac{1}{h} \p{\int_1^{1+h} f\p{x} \,\diff{x} - \int_0^h f\p{x} \,\diff{x}} \\ &\leq \liminf_{h\to0}\,\frac{1}{h} \p{hf\p{1 + h} - hf\p{0}} \\ &= \liminf_{h\to0}\,\frac{1}{h} \p{hf\p{1} - hf\p{0}} \\ &= f\p{1} - f\p{0}, \end{aligned}$$

   as required.

2. Notice that because all the $f_n$ are non-decreasing, it follows that $F$ is non-decreasing as well. Thus, $F$ is differentiable almost everywhere. Similarly, $r_n$ is differentiable almost everywhere as well, for all $n \geq 1$.

   By assumption, $F$ converges everywhere so $r_n\p{x} \to 0$ for all $x \in \br{0, 1}$. We also have

   $$\begin{aligned} F\p{x} &= \sum_{k=1}^n f_k\p{x} + r_n\p{x} \\ \implies F'\p{x} &= \sum_{k=1}^n f_k'\p{x} + r_n'\p{x} \end{aligned}$$

   almost everywhere since all three functions are differentiable almost everywhere. Thus, for almost every $x \in \br{0, 1}$,

   $$r_n'\p{x} - r_{n+1}'\p{x} = f_n'\p{x} \geq 0,$$

   since $f_n$ is non-decreasing. In other words, $\set{r_n'\p{x}}_n$ and because each $f_n$ is non-decreasing,

   $$\sum_{k=n}^m f_n'\p{x} \geq 0\ \forall m \implies r_n'\p{x} \geq 0,$$

   so $r_n'\p{x}$ is a monotone decreasing sequence bounded below. Thus, $r_n'$ converges almost everywhere, so we may apply the monotone convergence theorem. Combined with (1), we have

   $$\int_0^1 \lim_{n\to\infty} r_n'\p{x} \,\diff{x} = \lim_{n\to\infty} \int_0^1 r_n'\p{x} \,\diff{x} \leq \lim_{n\to\infty} \p{r_n\p{1} - r_n\p{0}} = 0.$$

   Since $r_n' \geq 0$, it follows that $\lim_{n\to\infty} r_n'\p{x} = 0$ almost everywhere and so

   $$F'\p{x} = \sum_{k=1}^n f_k'\p{x} + r_n'\p{x} \xrightarrow{n\to\infty} \sum_{k=1}^\infty f_k'\p{x},$$

   as required.